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ch3-soln - — 31 Chapter 3 Problem Solutions 3.1 Determine...

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Unformatted text preview: — 31 Chapter 3 Problem Solutions 3.1 Determine the highest speed at which two generators mounted on the same shaft can be driven so that the frequency of one generator is 60 Hz and the frequency of the other is 25 Hz. How many poles does each machine have? Solution: Let P = number of poles: 2x60x60 2x60x25 speed = P60 = P25 {’29 _ E9 P25 ' 25 P60 and P25 must be even integral numbers lowest value where P50 = 2.41325. Thus, P25 = 10 P50 = 24 3.2 The three-phase synchronous generator described in Example 3.1 is operated at 3600 rpm and supplies a unity poWer factor load. If the terminal voltage of the machine is 22 kV and the field current is 2500 A, determine the line current and the total power consumptiOn of the load. Solution: Using the values in the solution of Example 3.1, 45855 _ at“ = 3838 x2000V _ 29869.1 V Given: VLL = 22 kV yam = (J2/J3’)x22000V=17962.9v If on = 17962.9 coswt, then in = in“: 0056.115 and ea! = 17962.9coswt —4.1484 x 10‘3 x 1207: x iamsinwt 17962.9 coswt — 1.563911.“ sin wt 17962.92 +(1.56391‘%“)2 = 29869.1 V 15259.4 A iam/fl = 10.79 kA x/Ex 22 x 10.79 x 1 MW = 411.2 MW m D II Hence, z"l a“ L: P3¢> 3.3 A three-phase round-rotor synchronous generator has negligible armature re- sistance and a synchronous reactance Xd of 1.65 per unit. The machine is connected directly to an infinite bus of voltage 104$ per unit. Find the in- ternal voltage E.- of the machine when it delivers a current of (a) 1.04 30° per 32 unit, (b) 1.04m per unit and (c) 1.04—30" per unit to the infinite bus. Draw phasor diagrams depicting the operation of the machine in each case. Solution: Ei/Lo = VaLgi+IaL6XdL_QOO = 1.04Q+1.0L9><1.65490° = 1.0&°+1.65590°+0 (a) 0 = 30° EiLé = 1.0&°+1.654120° 1.44183° per unit Va=1.0p.u. (b) 0 = 0° = 1.0fl+1.65490° E‘L5 t _ o . 1.65 p.u. = 1.934 58.8 per umt 6 = -30° Ei[_6 1.0&+1.65460° = 2.324 38° per unit 33 3.4 A three-phase round-rotor synchronous generator, rated 10 kV, 50 MVA has armature resistance R of 0.1 per unit and synchronous reactance Xd of 1.65 per unit. The machine operates on a 10 kV infinite bus delivering 2000 A at 0.9 power factor leading. (a) Determine the internal voltage Bi and the power angle 6 of the machine. Draw a phasor diagram depicting its operation. (b) What is the open-circuit voltage of the machine at the same level of exci- tation? (c) What is the steady—state short-circuit current at the same level of excita- tion? Neglect all saturation effects. Solution: (a) Choosing Vb = 10 kV and MVAb = 50 MVA: Z, = (0.1+ j1.65) per unit = 1.653(8653" 2 244g 0 V = l—O per unit = 1.0 per unit 50 x 103 I = —— A = 2886.75A b «E x 10 2000 . . Ia. — 2886.75 per unlt — 0.693 per unit 9 = 005'1 0.9 = 258" Big — VzL°+IaZaza+0 1.0L0'1-l- 0.693 x 1.6530; 112.37” per unit = 1.195; 61.83° = 11.9516183‘j kV (b) Open-circuit voltage: E,- = 11.95 kV (c) Short-circuit voltage: E1- 1.195 E 1.653 per unit = 0.7242 per unit = 2090.7 A in; "‘2“ ~ 1 «a .;; «E ,5! n w 3%»: :-gx‘.~a’s'zz‘ 125%”:3‘; V_;r:':;:'_;::; f1" .’ 15:: g ; 2 3.13:“:52: 1; a: :51; "r W: .gr 4-4:“ a v 5:} ta 4%. v gel—3% ‘ a—J-s—sz: Esq—m: sail .» "Fa-2,. .. ca: 445“.» 34 3.5 A three-phase round-rotor synchronous generator, rated 16 kV and 200 MVA, has negligible losses and synchronous reactance of 1.65 per unit. It is operated on an infinite bus having a voltage of 15 kV. The internal emf E,- and the power angle 6 of the machine are found to be 24 kV (line-to—line) and 274°, respectively. ((1) Determine the line current and the three-phase real and reactive power . being delivered to the system. (b) If the mechanical power input and the field current of the generator are now changed so that the line current of the machine is reduced by 25% at the power factor of (a), find the new internal emf E.- and the power angle 6. (c) While delivering the reduced line current of (b), the mechanical power input and the excitation are further adjusted so that the machine operates at unity power factor at its terminals. Calculate the new values of Bi and 6. Solution: (a) Using 16 kV, 200 MVA base; Va = 15/16 per unit = 09375 per unit _ EiLé = %4 274° per unit = 1.5127.4° per unit EiLé—Vafl = InXd490°—0 1.54 274° — 0.937540: = L, x 1.65; 90° — 9 In 4 90° — 0 = 0.4818160.27° per unit In 5-0 = 048184-2973" per unit 3 Base I = M kA = 7.217 kA \/§ x 16 Therefore, I a S 0.4818 x 7.217 kA = 3.477 kA 0.9375 x 0.4818 per unit = 0.4517 per unit 90.34 MVA Thus, P = 90.34 cos 29.73° MW = 78.45 MW Q = 90.34 sin 29.73° Mvar = 44.80 Mvar ll (5) New Ia. = 0.75 x 0.4818 per unit = 0.3614 per unit 90 -— 6 = 60.27° EiLb‘ = VGLO°+IQX4490°—6 0.9375¢°+ 0.3614 x 1.65 4 6027" = 1.3374 228° per unit 21.45228“ kV L-L 35 NewIa = 0.3614 per unit 9 = 0° Eiéé = 0.937540:+ 0.3614 x 1.654 90° 1.1114 325° per unit = 17.8; 325° kV .w 2:. § arm.“ . < 3.6 The three-phase synchronous generator of Prob. 3.5 is operated on an infinite bus of voltage 15 kV and delivers 100 MVA at 0.8 power factor lagging. (a) Determine the internal voltage Ei, power angle 6 and the line current of the machine. 3, (b) If the field current of the machine is reduced by 10%, while the mechanical power input to the machine is maintained constant, determine the new value of 6 and the reactive power delivered to the system. (c) The prime mover power is next adjusted without changing the excitation so that the machine delivers zero reactive power to the system. Determine the new power angle 6 and the real power being delivered to the system. ., was. 1%; «H w ”again A; (d) What is the maximum reactive power that the machine can deliver if the level of excitation is maintained as in (b) and (c)? Draw a phasor diagram for the operation of the machine in cases (a), (b) and (c). : Solution: ‘ . (a) From Prob. 3.5, Va(or V,) = 0.9375 per unit S = 0.5 per unit Xd = 1.65 per unit 0 = -36.9° IO = S/Va = 0.5/0.9375 per unit E./_6 = VGL01+IEX¢¢90°+0 0.5 = 0.9375 0° . ' ° — . ° ' L + 0.9375 x 1 6o 5 90 36 9 per unit 1.6258125]o per unit = 26.0; 25.7" kV 36 New E,- 0.9 x 1.6258 per unit = 1.46322 per unit P = 0.5 x 0.8 per unit 04 per unit . PXd ‘ . _ 0.4x 1.65 — _1 -— = 1 ——————— = o 6 " 5m (1413.) 51“ (0.9375x1.46322) 2876 V X—d— ——(E- c056— Vt) 0.9375 II New Q (1.46322 cos 2876" —— 0.9375) 0.196 per unit = 39.2 Mvar (c) When Q = 0, V 0.9375 — —1 —l = _1 = — o 6 —— cos E1- cos (1.46322) 00.15 EV 0.9375 1.463 2 P = —XT sin6= —1—:35—2 sin 5015" per unit = 0.638 per unit 127.65 MW (d) For V}, E. and Xd fixed, Qmu occurs when 6 = 0°. Hence, V, 0.9375 ma = E' —V = . —0.9 7— ' Q 2 Xd ( t) 1.65 (1 46322 3 a) per umt = 0.2787 per unit = 59.74 Mvar =1.626 p.u. v.=o.9375 p.11. 3.7 Starting with Eq. (3.31), modify Eq. (3.38) to show that lVl P = W{IEI(Rcosé+dein6)— thR} V Q = R.“'X2{Xd<IE|cos6Iv.|1—R|E.|sm5} when the synchronous generator has non-zero armature resistance R. 37_ Solution: From Eq. (365), V = E + Ia(R +de) and, therefore, lEilLé—thldfi Ia = . (R'l'JXd) , I. : IEild—thl “ (R-J'Xd) 3 . lellEild— W ‘1 S = P+ = VI. = ————-.—-— i ’Q ’ “ (12—an = MHEz-Hcosé-jsinél- W (R-J'Xd) Z _ lellEiHcosé—jsiMMR-i—J’Xd) — W (212+an l _ (R2 +X3) : Separating real and imaginary parts, 1 thllEx‘l . M121? ; P (R2 +X3){ cos +dem } (R2+Xg) : V = Ez—li—lfg {IEil (Rcos6 + deiné) -— [th R} d _ thllEil _ . Mlsz Q _. (R2 +X3) {Xdcosé Rsm6} ——(R2 +X3) lel m {X4 (IEilCOS5 - thl) — R lEilSm 5} 3.8 The three-phase synchronous generator described in Example 3.4 is now oper— ated on a 25.2 kV infinite bus. It is found that the internal voltage magnitude lEil = 49.5 kV and that the power angle 6 = 38.5". Using the loading capa- bility diagram of Fig. 3.14, determine graphically the real and reactive power delivered to the system by the machine. Verify your answers using Eqs. (3.38). Solution: 25.2 |V,| = 25.2 IN = W per unit = 1.05 per unit 49.5 . ‘ . lEi[ = 49.5 kV = 3— per unit = 2.0625 per unit The distance 72—]: on the loading capability diagram is IEil 2.0625 “4le = ——1‘05 x 1.7241 units = 1.1393 units The angle formed by points k-n-o is 38.5". Hence, point I: is marked as shown. By reading from the chart, Pk = 0.7 per unit and QC = 0.31 per unit. P+jQ = IthQIPk+ij1 = 1.052(0.7+j0.31) per unit = 1.052 (0.7+ 30.31) x 635 MVA P = 490 MW Q = 217 Mvar 38 From Eq. (3.38), _ mm. P — Xd sxné 1.05x2.0625 , c, _ , = mess ) x635MW - 496.5 MW V Q = LX§<IE.icos6—Ivzl) 1.05 (2.0625cos(38.5°) —- 1.05) x 635 Mvar = 218.2 Mvar 1.7241 3.9 A three-phase salient-pole synchronous generator with negligible armature re- sistance has the following values for the inductance parameters specified in Table 3.1, , L3 = 2.7656 mH Mf = 31.6950 mH Ln = 0.3771 mH M3 = 1.3828 mH Lf; = 433.6569 mH During balanced steady—state operation the field current and a-phase armature current of the machine have the respective values if = 4000 A in = 20, 000 sin (0d — 30°) A (a) Using Eq. (3.41), determine the instantaneous values of the flux linkages Aa, Ab, Ac and A; when 6.1 = 60°. (b) Using Park’s Transformation given by Eqs. (3.42) and (3.43), determine the instantaneous values of the flux linkages Ad, Ag and A0, and the currents id, 2}, and 2'0 when 1% = 60°. (c) Verify results using Eqs. (3.45) - (3.46) Solution: (11) From Table 3.1, Lao Lab Lac La _Ms —M_, Labc é Lac Lab Lac = _Ms L: —M_, Lac Lab Lac “My ‘Ms L: ‘ c0520.; —cos2(0¢ +30°) —cos2(9d + 150°) + L... —cos2(9d+30°) cos2(0d—120°) —cos2(0d—90°) —cos2(0d+ 150°) cos2(9d —90°) cos2(6d+ 120°) 2.7656 —1.3828 -l.3828 Lag,c = —-1.3828 2.7656 —1.3828 -1.3828 —1.3828 2.7656 cos 120° — cos 180° — cos 420° + 0.3771 -— cos 180° cos (—120°) — cos (—60°) mH — cos 420° — cos (—60°) cos 360° — '39 —1.0057 2.57705 -l.57135 -l.57135 —1.57135 3.1427 2.57705 -1.0057 —1.57135 mH La; cos 8d Labc,f é [ Lbf J = Mf COS (9d — 120°) LC, cos (6,, — 240°) cos 60° 15.8475 = 31.695 cos (—60°) mH = 15.8475 mH cos (—-_180°) ~31.695 lll> i,z sin (30°) 10 2'an 1;, = 20000 sin (—90°) A = -20 RA ic sin (—210°) 10 With 2', = 4 kA and L f f = 433.6569 mH, )‘a Lao Lab Lac Laf in Ab _ Lba Lbb Lbc Lbf i6 Ac — Lea ch Lac ch 1‘: A! Lfa Lfb Lfc Ln if 2.57705 -—1.0057 —1.57135 15.8475 10 _ -—1.0057 2.57705 —1.57135 15.8475 —20 Wb—T -_ —-1.57135 -1.57135 3.1427 —31.6950 10 15.8475 15.8475 —31.6950 433.6569 4 93.5610 -13.9215 = —79.6395 WM“ 12592026 (b) When ed = 60°, 1 1 _1 1 -1_ 1/3 2 2 2 7: ¢§ 3 = _ 6/3 _fi = 1 _1 P 3 2 2 0 75 72' 0 L 1 1 1 1 1 v5 7? 75‘ 3 75 75 1 1 2 Ad 73 73 1/; 93.5610 97.5381 = 1 1 = A., 75 —75 0 .13.9215 76.0016 WbT 1 1 1 A0 75 75 75 —79.6395 0 W P A«be - 1 1 2 ' u 73 7-5 1/; 10 —12.2474 - = 1 _ 1 = zq 75 75 0 —20 21.2132 kA ' 1 1 1 1,0 75 E 7—5 10 0 P “W 1nab: 40 L, = L3+M5—;Lm = 3.58275 mH @111, = 38.8183 mH ‘ A4 = Ldid+¢§Mm = 4.71405 x (—12.2474)+38.8183 x4 WbT = 97.5381 Wb-T Ag = Lqiq = 3.58275x21.2132 WbT = 76.0016 Wb-T A0 = Loio = 0 (since 10:0) 1, = fiMfid+me = 38.8183 x (—12.2474) +433.6569 x 4 war 1259.20 Wb-T 3.10 The armature of a three—phase salient-pole generator carries the currents z", = «5 x 1000 sin (81 — 0a) A 2'2, = \/§ x 1000 sin (6d — 120° — ea) A «5 x 1000 sin (0.: — 240° — an) A a. n l (0) Using the P-Transformation matrix of Eq. (3.42), find the direct-axis cur- rent id and the quadrature-axis current iq. What is the zero.sequence cur— rent 1'0? (b) Suppose that the armature currents are 1;, = 1/2' x1000sin(0d—0a) A 2'1, = ic = 0 Determine id, 1", and 1'0. Solution: (a) id 20 1}, = P ‘Lb to re 1 1 . = 3 15: 1c: -(1) x mom/5 sfn (9d — 9a) 8 3 2 “2 5m(9a—0a—12O) A 712. 715 715 sin (0d — 6,, — 240°) H g E Q. :1 Q) n. I Q: 0 + W o O v t» 41 1 1d 2 § 5 —% sin(0d — 90) - -1. 1 .1. 0 10 J2 E «2' é—sin (6d — 9a) 2000 \/§ 2 5111(041 a) i sin (9d — 05) 3.11 Calculate the direct-axis synchronous reactance Xd, the direct-axis transient reactance X [1 and the direct-axis subtransient reactance X {1’ of the 60 Hz' salient- pole synchronous machine with the following parameters: L5 = 2.7656 mH Lff = 433.6569 mH Ln = 4.2898 mH M5 = 1.3828 mH Mf = 31.6950 mH MD = 3.1523 mH Ln 2 0.3771 mH M, = 37.0281 mH Solution: 3 , 3 Ld = L + Ms — 5L," = 2.7606+ 1.3828 — 5 x 0.3771 mH = 4.71405 mH X1 = 1207.— x 4.71405 x 10—39 = 1.777 n , 3 M} _ 3 31.69502 Ld — Ld — E f}? — 4.71400—5 X 433.6569 mH — 1.2393 mH x; = 12077 x 1.2393 x 10—30 = 0.467 n L” L 3 MfLD'l'MIQDLff—szMDM, d ‘ “'5 LffLD—ME _ 4 71405 _3 316952 x 4.2898 + 3.15232 x 433.6569 — 2 x 31.6950 x 3.1523 x 37.0281 ' 2 433.6569 x 4.2898 — 37.02812 0.9748 mH X}; = 12077 x 0.9748 x 10-39 = 0.367 n 3.12 The single~1ine diagram of an unloaded power system is shown in Fig. 3.22. Reactances of the two sections of transmission line are shown on the diagram. The generators and transformers are rated as follows: ) 1nH . .mwmci‘mfiu‘nm': :; 42 Generator 1: 20 MVA, 13.8 IN, ’1; = 0.20 per unit Generator 2: 30 MVA, 18 kV, Xé’ = 0.20 per unit Generator 3: 30 MVA, 20 kV, X (’1’ = 0.20 per unit Transformer T1: 25 MVA, 220Y/13.8A kV, X = 10% Transformer T 2: Single-phase units each rated 10 MVA,127/ 18 kV, X = 10% Transformer T3: 35 MVA, 220Y/22Y kV, X = 10% ((1) Draw the impedance diagram with all reactances marked in per unit and with letters to indicate points corresponding to the single-line diagram. Choose a base of 50 MVA, 13.8 kV in the circuit of Generator 1. (b) Suppose that the system is unloaded and that the voltage throughout the system is 1.0 per unit on bases chosen in part (a). If a three-phase short cir— cuit occurs from bus C to ground, find the phasor value of the short-circuit current (in amperes) if each generator is represented by its subtransient reactance. (c) Find the megavoltamperes supplied by each generator under the conditions of part (b). Solution: (a) 50.20 j0.0826 C j0.1033 j0.167 Gen 1: X” = 0.2 x 553- = 0.50 per unit 391) rating T2 = 220/ 18 kV, 30 MVA Base in trans. line: 220 kV, 50 MVA Base for Gen 2 = 18 kV Gen 2: X” = 0.2 x g = 0.333 per unit Base for Gen 3 = 22 kV 20 2 50 Ge 1 H = . — — = . . n3 X 02(22) x 30 0275perun1t 50 . Transformer T1: X = .01 x g = 0.20 per unit Transformer T2: X = .01 x 35% = 0.167 per unit Transformer T3: X = .01 x 3—: = 0.143 per unit Transmission lines: 2 Base Z = 2?) = 968 Q 00 80 100 _ = . ' — = 0. O ' 968 0 0826 per umt 968 1 33 per umt SIC 0.7826 per unit 0.6033 per unit = 0.418 per unit ll per unit = 1.2785—90° per unit 1.658 5—90° per unit 7 = 0—41—85—900 per unit = 2.392(—-90° per unit .7 3 . ' Eh = E,- = E,‘ = 1.0[_0: per unit X1 —— 0.50 + 0.20 + 0.0826 per unit X2 — 0.333 + 0.167 + 0.1033 per unit X3 = 0.143 + 0.275 per unit E 1 c, 11 ’ 3J7; ’ 0.78261399— I - E1 - 1 —90° er unit 2 _ jXQ _ 0.6033;- F 13 = E 1 If = I1 + I2 + I3 = 50 x 106 IfieatC = -—————A = 131.22A b J27 x 220 x 103 . |If| = 5.328 x 131.22 A = 699 A 1521 = Eing = 1.0x 1.658 x 50 MVA = 82.9 MVA {53] = EisIg = 1.0 x 2.392 X 50 MVA = 119.6 MVA 3.13 The ratings of the generators, motors and transformers of Fig. 3.23 are: Generator 1: Generator 2: Synchronous motor 3: Three-phase Y—Y transformers: Three-phase Y—A transformers: 20 MVA, 18 IN, 1;: 20% 20 MVA, 18 kV, X” = 20% 30 MVA, 13.8 kV, X” = 20% 20 MVA, 138Y/20Y kV, X = 10% 15 MVA, 138Y/13.8A kV, X = 10% 43 (1.278 + 1.658 + 2.392) (—90° per unit = 5.328 (490° per unit 44 (0) Draw the impedance diagram for the power system. Mark impedances in per unit. Neglect resistance and use a base of 50 MVA, 138 kV in the 40-9 line. v (1)) Suppose that the system is unloaded and that the voltage throughout the system is 1.0 per unit on bases chosen in part (a). If a three-phase short cir- cuit occurs from bus C to ground, find the phasor value of the short-circuit current (in amperes) if each generator is represented by its subtransient reactance. (c) Find the megavoltamperes supplied by each synchronous machine under the conditions of part (b). Solution: (0) Base voltages are: 40 {2 lines 138 kV 20 (2 lines 138 kV Gen. 1 & 2 20 kV Motor 3 13.8 kV . . . 1382 Base 1mpedance in lines = 30— = 381 Q 40 40 Q 1' : = — = . 0 ‘ me Z 381 0 1 5 per unit . . _ 20 _ . 20 $2 hne. Z _. EST — 0.053 per umt Transformers: 50 Y-Y = 0.1 x 50 = 0.250 per unit , 50 . X-A = 0.1 x ~1—5 = 0.333 per umt Gens. 1 & 2: 2 18 50 XII = 0 _- — = 7 ~ 20 x (20) x 20 0.40:: per unit 45 Motor 3: 50 X" = 0.20 x E = 0.333 per unit (b) If a fault occurs at C, by symmetry equal currents are input from generators 1 and 2. Moreover, no current should exist between busses A and B through the 30.105 per unit branch. If this branch is omitted from the circuit, the system simplifies to S/C Eh - Eu = Eu = LOLOO. per unit X1 — X2 = 0.405+O.250 +0053 + 0.333 per unit = 1.041 per unit X3 — 0.333 per unit _ _ [Eill __ 1.0 . _ . [II] _ [12] _ 1X1} —. 1.041 per unit — 0.9606 per unit Bi 1.0 . . IIgI = I]X;:|| = 0.333 per unit = 3.0 per unit [Ifl = 4.9212 per unit 50 x 106 135 tC = —————A=2091.8A baa fix13.8x103 w = 4.9212 x 2091.8 A = 10.29 kA (C) 3511 = ngl = 1.0 x 50 x 0.9606 MVA = 48.03 MVA |33| = 1.0 x 50 x 3.0 MVA = 150 MVA Chapter 4 Problem Solutions 4.1 The all-alumin conductor identified by the code word Bluebell is composed of 37 strands ,ch having a diameter of 0.1672 in. Tables of characteristics of all—aluminu conductors list an area of 1,033,500 cmil for this conductor (1 cmil = (7/4) x 10‘6 in2). Are these values consistent with each other? Find the overall are of the strands in square millimeters. Solution: diameter = 0.1672x 1000 = 167.2 mils/strand ...
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